Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 286: 32


$\frac{3\sqrt 3+2\pi}{6}$

Work Step by Step

Given: $arccos({y-\frac{\pi}{3}})=\frac{\pi}{6}$ $({y-\frac{\pi}{3}})=cos(\frac{\pi}{6})$ Apply definition of cosine. $({y-\frac{\pi}{3}})=\frac{\sqrt 3}{2}$ $y=\frac{\sqrt 3}{2}+\frac{\pi}{3}$ $y=\frac{3\sqrt 3+2\pi}{6}$ Hence, $y=\frac{3\sqrt 3+2\pi}{6}$
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